Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > sci.physics > #690702 > unrolled thread

Shooting to the Moon

Back to article view | Back to sci.physics

Contents

  Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 05:20 -0700
    Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-15 08:55 -0400
      Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-15 08:34 -0700
        Re: Shooting to the Moon Sergio <invalid@invalid.com> - 2018-06-15 13:42 -0500
          Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 14:26 -0700
          Re: Shooting to the Moon benj <benj@nobody.net> - 2018-06-15 19:17 -0400
            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 18:21 -0700
              Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 05:55 -0700
              Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-17 17:04 -0700
                Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 18:13 -0700
                  Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-17 18:44 -0700
                    Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 19:07 -0700
                      Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-18 13:35 -0700
                        Re: Shooting to the Moon Odd Bodkin <bodkinodd@gmail.com> - 2018-06-18 21:02 +0000
                          Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-19 10:39 -0700
                            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-19 15:35 -0700
                              Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-19 21:09 -0700
                              Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-20 10:49 -0700
                                Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-20 16:18 -0700
                                  Re: Shooting to the Moon Edward Prochak <edprochak@gmail.com> - 2018-06-21 15:05 -0700
                                    Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-21 16:57 -0700
                        Re: Shooting to the Moon Sergio <invalid@invalid.com> - 2018-06-18 16:18 -0500
                          Re: Shooting to the Moon Steve Carroll <frelwizzen@gmail.com> - 2018-06-19 10:40 -0700
                          Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-19 15:20 -0700
                        Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-18 22:40 -0700
        Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-15 14:29 -0700
          Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-16 11:10 -0400
            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 15:13 -0700
            Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 19:04 -0700
              Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 20:05 -0700
                Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-16 21:18 -0700
                Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-17 05:35 -0400
                  Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 03:08 -0700
                    Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 06:00 -0700
                      Re: Shooting to the Moon Wally W. <ww84wa@aim.com> - 2018-06-17 13:28 -0400
                        Re: Shooting to the Moon Arindam Banerjee <banerjeeadda1234@gmail.com> - 2018-06-17 16:15 -0700
      Re: Shooting to the Moon benj <benj@nobody.net> - 2018-06-15 11:48 -0400
    Re: Shooting to the Moon numbernumber1964@gmail.com - 2018-06-18 15:29 -0700
    Re: Shooting to the Moon numbernumber1964@gmail.com - 2018-06-19 12:21 -0700
      Re: Shooting to the Moon Odd Bodkin <bodkinodd@gmail.com> - 2018-06-19 21:47 +0000
    Re: Shooting to the Moon numbernumber1964@gmail.com - 2018-06-19 12:29 -0700
    Re: Shooting to the Moon furthermore7990662@gmail.com - 2018-06-19 16:03 -0700
    Re: Shooting to the Moon furthermore7990662@gmail.com - 2018-06-19 16:03 -0700
    Re: Shooting to the Moon carleto4157990662@gmail.com - 2018-06-19 16:05 -0700
    Re: Shooting to the Moon carleto4157990662@gmail.com - 2018-06-19 16:06 -0700

Page 1 of 3  [1] 2 3  Next page →

#690702 — Shooting to the Moon

To send a 400Kg object to the Moon, you need a 1Km long rail gun (of my new, low voltage design) using 25 million average amperes of current on the rails.  

Quite doable. 

The exit velocity should be about 55,000m/s which would mean  a lot of acceleration. What will happen if some lifeform is put in it? I don't think it will get harmed, going by the experience of the high rise lifts that use internal force linear motors for great acceleration, as opposed to the kind of pull-up external force stuff that Einstein bunglingly used to justify the notions of GR. 

would , 
Arindam Banerjee 

[toc] | [next] | [standalone]

#690707

On Fri, 15 Jun 2018 05:20:36 -0700 (PDT), Arindam Banerjee wrote:

>To send a 400Kg object to the Moon, you need a 1Km long rail gun (of my new, low voltage design) using 25 million average amperes of current on the rails.  
>
>Quite doable. 
>
>The exit velocity should be about 55,000m/s which would mean  a lot of acceleration. What will happen if some lifeform is put in it? I don't think it will get harmed, going by the experience of the high rise lifts that use internal force linear motors for great acceleration, as opposed to the kind of pull-up external force stuff that Einstein bunglingly used to justify the notions of GR. 
>
>would , 
>Arindam Banerjee 

Do you have values for:
1. The Mach number at your exit velocity?
2. The acceleration, assuming constant acceleration ... over a
distance of 1 km ... from zero initial velocity?

[toc] | [prev] | [next] | [standalone]

#690715

On Friday, June 15, 2018 at 8:55:34 AM UTC-4, Wally W. wrote:
> On Fri, 15 Jun 2018 05:20:36 -0700 (PDT), Arindam Banerjee wrote:
> 
> >To send a 400Kg object to the Moon, you need a 1Km long rail gun (of my new, low voltage design) using 25 million average amperes of current on the rails.  
> >
> >Quite doable. 
> >
> >The exit velocity should be about 55,000m/s which would mean  a lot of acceleration. What will happen if some lifeform is put in it? I don't think it will get harmed, going by the experience of the high rise lifts that use internal force linear motors for great acceleration, as opposed to the kind of pull-up external force stuff that Einstein bunglingly used to justify the notions of GR. 
> >
> >would , 
> >Arindam Banerjee 
> 
> Do you have values for:
> 1. The Mach number at your exit velocity?

This can't be right:  Mach 160?

> 2. The acceleration, assuming constant acceleration ... over a
> distance of 1 km ... from zero initial velocity?

Ed

[toc] | [prev] | [next] | [standalone]

#690737

On 6/15/2018 10:34 AM, Edward Prochak wrote:
> On Friday, June 15, 2018 at 8:55:34 AM UTC-4, Wally W. wrote:
>> On Fri, 15 Jun 2018 05:20:36 -0700 (PDT), Arindam Banerjee wrote:
>>
>>> To send a 400Kg object to the Moon, you need a 1Km long rail gun (of my new, low voltage design) using 25 million average amperes of current on the rails.  
>>>
>>> Quite doable. 
>>>
>>> The exit velocity should be about 55,000m/s which would mean  a lot of acceleration. What will happen if some lifeform is put in it? I don't think it will get harmed, going by the experience of the high rise lifts that use internal force linear motors for great acceleration, as opposed to the kind of pull-up external force stuff that Einstein bunglingly used to justify the notions of GR. 
>>>
>>> would , 
>>> Arindam Banerjee 
>>
>> Do you have values for:
>> 1. The Mach number at your exit velocity?
> 
> This can't be right:  Mach 160?
> 
>> 2. The acceleration, assuming constant acceleration ... over a
>> distance of 1 km ... from zero initial velocity?
> 
> Ed
> 

problem is getting 25,000,000 amps through 2 sliding connections for
1km.   the Navy gave up.

[toc] | [prev] | [next] | [standalone]

#690763

Never say die.

[toc] | [prev] | [next] | [standalone]

#690782

On 6/15/2018 2:42 PM, Sergio wrote:
> On 6/15/2018 10:34 AM, Edward Prochak wrote:
>> On Friday, June 15, 2018 at 8:55:34 AM UTC-4, Wally W. wrote:
>>> On Fri, 15 Jun 2018 05:20:36 -0700 (PDT), Arindam Banerjee wrote:
>>>
>>>> To send a 400Kg object to the Moon, you need a 1Km long rail gun (of my new, low voltage design) using 25 million average amperes of current on the rails.
>>>>
>>>> Quite doable.
>>>>
>>>> The exit velocity should be about 55,000m/s which would mean  a lot of acceleration. What will happen if some lifeform is put in it? I don't think it will get harmed, going by the experience of the high rise lifts that use internal force linear motors for great acceleration, as opposed to the kind of pull-up external force stuff that Einstein bunglingly used to justify the notions of GR.
>>>>
>>>> would ,
>>>> Arindam Banerjee
>>>
>>> Do you have values for:
>>> 1. The Mach number at your exit velocity?
>>
>> This can't be right:  Mach 160?
>>
>>> 2. The acceleration, assuming constant acceleration ... over a
>>> distance of 1 km ... from zero initial velocity?
>>
>> Ed
>>
> 
> problem is getting 25,000,000 amps through 2 sliding connections for
> 1km.   the Navy gave up.
> 
Lucky that when you are an "idea man" you don't need to solve problems 
like 25,000,000 amps through a connector. You just give everyone your 
idea and someone else works out the details!

[toc] | [prev] | [next] | [standalone]

#690798

Did 3200 amps with own two brown hands 

[toc] | [prev] | [next] | [standalone]

#690839

On Saturday, June 16, 2018 at 11:21:25 AM UTC+10, Arindam Banerjee wrote:
> Did 3200 amps with own two brown hands

25 million amps is nothing but 25000 volts divided by 1 milliohm. 

No problem at all, both in theory and practice.

I got 3200 amps by dividing 16V with 5 milliohms, the contact resistance for my new design low voltage low resistance rail gun.

As we can see in the following video films:

https://www.youtube.com/watch?v=hqBfwAClVlg
IFE - 1 Ground Experiments

https://www.youtube.com/watch?v=w9eGq4Oiv9s
IFE - 2 Experimental setups

https://www.youtube.com/watch?v=V3hC48BMrno
IFE - 3 Pendulum experiments

https://www.youtube.com/watch?v=3sSPxGsLkws
IFE - 4 Evolution of spaceship

https://www.youtube.com/watch?v=pJdM6UDPauU
IFE - 5 Hydrogen Transmission Network

https://www.youtube.com/watch?v=TUAcx7rAplc
IFE - 6 Spaceship Design

https://www.youtube.com/watch?v=H5Zbpvc3fdA
IFE - 7 Anti-Gravity

https://www.youtube.com/watch?v=VA9LUwqMhxY
IFE - 8 New Physics

Cheers,
Arindam Banerjee

[toc] | [prev] | [next] | [standalone]

#691001

On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> Did 3200 amps with own two brown hands

you are off by 3 orders of magnitudes!

[toc] | [prev] | [next] | [standalone]

#691011

On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > Did 3200 amps with own two brown hands
> 
> you are off by 3 orders of magnitudes!

No.

I shorted with a 16.5 V 250F capacitor bank with a low milliohm circuit resistance to get 3200 amps on the average.  The 3200 A was calculated from the drop in the voltage across the capacitor bank. Say when it dropped from 16.5 V to 7.5 V then the charge discharged is 250 (16.5 - 7.5) = 2250 coulombs.

Agreed?

Now this discharge takes place as long as there is contact between the armature and the rails. This is measured as say 0.7 seconds

Now by defintion of current, current is charge passed through a circuit within a given time. Or, i = Q/t

Thus i = 2250/0.7 = 3210 Amps (average)

This is the current which gave the armature the greater exit velocity to the rail gun armature as compared to the earlier case when there was no current.

Thus proving:
1. that the gun does work - electricity makes the armature move
2. when mounted as a pendulum there is no extra swingback for the current case meaning that the Lorenz force (should be called Thomson-Heaviside force or THE force) has no reaction. This is a stunningly important breakthrough in the field of machanics
3. When the armature is arrested the whole system moves forward thus violating Newton's first law of motion, as indeed has to be the case when the third law is violated as per the pendulum experiment.

The following video films make it all very clear:

https://www.youtube.com/watch?v=hqBfwAClVlg
IFE - 1 Ground Experiments

https://www.youtube.com/watch?v=w9eGq4Oiv9s
IFE - 2 Experimental setups

https://www.youtube.com/watch?v=V3hC48BMrno
IFE - 3 Pendulum experiments

https://www.youtube.com/watch?v=3sSPxGsLkws
IFE - 4 Evolution of spaceship

https://www.youtube.com/watch?v=pJdM6UDPauU
IFE - 5 Hydrogen Transmission Network

https://www.youtube.com/watch?v=TUAcx7rAplc
IFE - 6 Spaceship Design

https://www.youtube.com/watch?v=H5Zbpvc3fdA
IFE - 7 Anti-Gravity

https://www.youtube.com/watch?v=VA9LUwqMhxY
IFE - 8 New Physics

One has to spend several days to go through all of these new discoveries.

Cheers,
Arindam Banerjee

[toc] | [prev] | [next] | [standalone]

#691018

On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> > On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > > Did 3200 amps with own two brown hands
> > 
> > you are off by 3 orders of magnitudes!
> 
> No.
> 
> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> circuit resistance to get 3200 amps on the average.  The 3200 A
> was calculated from the drop in the voltage across the capacitor
> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> 
> Agreed?

Yes, but
That is still 3 orders of magnitude less than the
25,000,000 Amps needed for the 1km rail gun.
I don't know what materials will carry that high a current.

ED

[toc] | [prev] | [next] | [standalone]

#691028

On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> > On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> > > On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > > > Did 3200 amps with own two brown hands
> > > 
> > > you are off by 3 orders of magnitudes!
> > 
> > No.
> > 
> > I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> > circuit resistance to get 3200 amps on the average.  The 3200 A
> > was calculated from the drop in the voltage across the capacitor
> > bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> > discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> > 
> > Agreed?
> 
> Yes, but
> That is still 3 orders of magnitude less than the
> 25,000,000 Amps needed for the 1km rail gun.
> I don't know what materials will carry that high a current.
> 
> ED

In my PhD work, my guide asked me to do a worksheet showing how much a conductor of a given dimension will heat up over time with current.

That I did, and with that help I designed my 3200A gun. It does not become hot by the way, thanks to the dimensioning.

Extending that to my motor (the IFE) I worked out that with a 500mm*500mm copper cross section, the heat increase would be manageable for 3 million amperes.

I can use the same worksheet to work out the dimensions for 25million amps.

My guess is that it would be around 2000mm*1000mm.

That is a lot of copper, for 2000m rails, but it will carry 25 million amps. Note, this are rough figures. We also need to know that the whole barrel will be vacuum sealed and as cold as possible. Colder the temperature, better the conductivity, lower the contact resistance, etc. I am not talking about superconductivity, just higher conductivity with lower temperature.

I am giving over-the-top back of the envelope estimates. Naturally, when we get down to the calculas and do the fine slicing, the actual figures for a given design will emerge. Reducing the contact resistance is the key. 

Another point - we need existing defunct mineshafts going down 1-2Km for this gun. No point creating a 1000 m vertical structure, what! Natural side-support, in a hole!

I think we can with a limited budget make a gun that can launch small satellites into space. It can compete with rocket launches. 

Cheers,
Arindam Banerjee

[toc] | [prev] | [next] | [standalone]

#691104

On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> > On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> > > On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> > > > On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > > > > Did 3200 amps with own two brown hands
> > > > 
> > > > you are off by 3 orders of magnitudes!
> > > 
> > > No.
> > > 
> > > I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> > > circuit resistance to get 3200 amps on the average.  The 3200 A
> > > was calculated from the drop in the voltage across the capacitor
> > > bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> > > discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> > > 
> > > Agreed?
> > 
> > Yes, but
> > That is still 3 orders of magnitude less than the
> > 25,000,000 Amps needed for the 1km rail gun.
> > I don't know what materials will carry that high a current.
> > 
> > ED
> 
> In my PhD work, my guide asked me to do a worksheet showing
> how much a conductor of a given dimension will heat up over
> time with current.
> 
> That I did, and with that help I designed my 3200A gun. It does not become hot by the way, thanks to the dimensioning.
> 
> Extending that to my motor (the IFE) I worked out that with a 500mm*500mm copper cross section, the heat increase would be manageable for 3 million amperes.
> 
> I can use the same worksheet to work out the dimensions for 25million amps.
> 
> My guess is that it would be around 2000mm*1000mm.
> 
> That is a lot of copper, for 2000m rails, but it will carry 25 million amps. Note, this are rough figures. We also need to know that the whole barrel will be vacuum sealed and as cold as possible. Colder the temperature, better the conductivity, lower the contact resistance, etc. I am not talking about superconductivity, just higher conductivity with lower temperature.
> 
> I am giving over-the-top back of the envelope estimates. Naturally, when we get down to the calculas and do the fine slicing, the actual figures for a given design will emerge. Reducing the contact resistance is the key. 
> 
> Another point - we need existing defunct mineshafts going down 1-2Km for this gun. No point creating a 1000 m vertical structure, what! Natural side-support, in a hole!
> 
> I think we can with a limited budget make a gun that can launch small satellites into space. It can compete with rocket launches. 
> 
> Cheers,
> Arindam Banerjee

Earlier you Arindam) posted:
"25 million amps is nothing but 25000 volts divided by 1 milliohm. 
No problem at all, both in theory and practice."

the total resistance needs to be 1 milliohm (or less)

So what is the resistance?
to obtain a voltage drop of 25000V over 2 kilometers,
R=rl/A
l - length 2 km (=2000 m)
A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
r - rho resistivity for copper is 0.000 000 017 ohm-meter

R= 0.000 017 ohm = 0.017 milliohm.

I'm impressed. about a factor of 5 margin of safety.

Now 2 Km by 2m^2 is a volume of 4000 m^3
Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
total weight 35,840,000 kg
Now I'm impressed again. Cost is a little over $25 million.

Of course it will be the assembly costs that swamp the material costs.
You'd likely want the rails to be continuous. so you'd want to caste the rails in place maybe?Or can it be multiple parallel sheets of copper?
(how big is a roll of copper 1km long?)

Physically doable, though challenging. that the navy gave up on their
railgun, suggests there are other challenges we haven't identified.

Nothing better be around the launch tip, that is going to be
one hell of a sonic boom!

Ed
Meaning the resistance needs to be 

[toc] | [prev] | [next] | [standalone]

#691110

Edward Prochak <edprochak@gmail.com> wrote:
> On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
>> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
>>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
>>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
>>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
>>>>>> Did 3200 amps with own two brown hands
>>>>> 
>>>>> you are off by 3 orders of magnitudes!
>>>> 
>>>> No.
>>>> 
>>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
>>>> circuit resistance to get 3200 amps on the average.  The 3200 A
>>>> was calculated from the drop in the voltage across the capacitor
>>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
>>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
>>>> 
>>>> Agreed?
>>> 
>>> Yes, but
>>> That is still 3 orders of magnitude less than the
>>> 25,000,000 Amps needed for the 1km rail gun.
>>> I don't know what materials will carry that high a current.
>>> 
>>> ED
>> 
>> In my PhD work, my guide asked me to do a worksheet showing
>> how much a conductor of a given dimension will heat up over
>> time with current.
>> 
>> That I did, and with that help I designed my 3200A gun. It does not
>> become hot by the way, thanks to the dimensioning.
>> 
>> Extending that to my motor (the IFE) I worked out that with a
>> 500mm*500mm copper cross section, the heat increase would be manageable
>> for 3 million amperes.
>> 
>> I can use the same worksheet to work out the dimensions for 25million amps.
>> 
>> My guess is that it would be around 2000mm*1000mm.
>> 
>> That is a lot of copper, for 2000m rails, but it will carry 25 million
>> amps. Note, this are rough figures. We also need to know that the whole
>> barrel will be vacuum sealed and as cold as possible. Colder the
>> temperature, better the conductivity, lower the contact resistance, etc.
>> I am not talking about superconductivity, just higher conductivity with lower temperature.
>> 
>> I am giving over-the-top back of the envelope estimates. Naturally, when
>> we get down to the calculas and do the fine slicing, the actual figures
>> for a given design will emerge. Reducing the contact resistance is the key. 
>> 
>> Another point - we need existing defunct mineshafts going down 1-2Km for
>> this gun. No point creating a 1000 m vertical structure, what! Natural
>> side-support, in a hole!
>> 
>> I think we can with a limited budget make a gun that can launch small
>> satellites into space. It can compete with rocket launches. 
>> 
>> Cheers,
>> Arindam Banerjee
> 
> Earlier you Arindam) posted:
> "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> No problem at all, both in theory and practice."
> 
> the total resistance needs to be 1 milliohm (or less)
> 
> So what is the resistance?
> to obtain a voltage drop of 25000V over 2 kilometers,
> R=rl/A
> l - length 2 km (=2000 m)
> A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> r - rho resistivity for copper is 0.000 000 017 ohm-meter
> 
> R= 0.000 017 ohm = 0.017 milliohm.
> 
> I'm impressed. about a factor of 5 margin of safety.
> 
> Now 2 Km by 2m^2 is a volume of 4000 m^3
> Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> total weight 35,840,000 kg
> Now I'm impressed again. Cost is a little over $25 million.
> 
> Of course it will be the assembly costs that swamp the material costs.
> You'd likely want the rails to be continuous. so you'd want to caste the
> rails in place maybe?Or can it be multiple parallel sheets of copper?
> (how big is a roll of copper 1km long?)
> 
> Physically doable, though challenging. that the navy gave up on their
> railgun, suggests there are other challenges we haven't identified.
> 
> Nothing better be around the launch tip, that is going to be
> one hell of a sonic boom!
> 
> Ed
> Meaning the resistance needs to be 
> 

I think the copper is going to melt the second you make a single firing.
Well, melt maybe is not the right word. Explosively vaporize is perhaps
better. Fairly simple to calculate the heating load and then ask whether
copper’s thermal conductivity is anywhere near what it would need to be. 

Quenching of accelerator superconducting dipoles is an object lesson. 

-- 
Odd Bodkin — Maker of fine toys, tools, tables

[toc] | [prev] | [next] | [standalone]

#691224

On Monday, June 18, 2018 at 5:02:57 PM UTC-4, Odd Bodkin wrote:
> Edward Prochak <edprochak@gmail.com> wrote:
> > On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> >> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> >>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> >>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> >>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> >>>>>> Did 3200 amps with own two brown hands
> >>>>> 
> >>>>> you are off by 3 orders of magnitudes!
> >>>> 
> >>>> No.
> >>>> 
> >>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> >>>> circuit resistance to get 3200 amps on the average.  The 3200 A
> >>>> was calculated from the drop in the voltage across the capacitor
> >>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> >>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> >>>> 
> >>>> Agreed?
> >>> 
> >>> Yes, but
> >>> That is still 3 orders of magnitude less than the
> >>> 25,000,000 Amps needed for the 1km rail gun.
> >>> I don't know what materials will carry that high a current.
> >>> 
> >>> ED
> >> 
> >> In my PhD work, my guide asked me to do a worksheet showing
> >> how much a conductor of a given dimension will heat up over
> >> time with current.
> >> 
> >> That I did, and with that help I designed my 3200A gun. It does not
> >> become hot by the way, thanks to the dimensioning.
> >> 
> >> Extending that to my motor (the IFE) I worked out that with a
> >> 500mm*500mm copper cross section, the heat increase would be manageable
> >> for 3 million amperes.
> >> 
> >> I can use the same worksheet to work out the dimensions for 25million amps.
> >> 
> >> My guess is that it would be around 2000mm*1000mm.
> >> 
> >> That is a lot of copper, for 2000m rails, but it will carry 25 million
> >> amps. Note, this are rough figures. We also need to know that the whole
> >> barrel will be vacuum sealed and as cold as possible. Colder the
> >> temperature, better the conductivity, lower the contact resistance, etc.
> >> I am not talking about superconductivity, just higher conductivity with lower temperature.
> >> 
> >> I am giving over-the-top back of the envelope estimates. Naturally, when
> >> we get down to the calculas and do the fine slicing, the actual figures
> >> for a given design will emerge. Reducing the contact resistance is the key. 
> >> 
> >> Another point - we need existing defunct mineshafts going down 1-2Km for
> >> this gun. No point creating a 1000 m vertical structure, what! Natural
> >> side-support, in a hole!
> >> 
> >> I think we can with a limited budget make a gun that can launch small
> >> satellites into space. It can compete with rocket launches. 
> >> 
> >> Cheers,
> >> Arindam Banerjee
> > 
> > Earlier you Arindam) posted:
> > "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> > No problem at all, both in theory and practice."
> > 
> > the total resistance needs to be 1 milliohm (or less)
> > 
> > So what is the resistance?
> > to obtain a voltage drop of 25000V over 2 kilometers,
> > R=rl/A
> > l - length 2 km (=2000 m)
> > A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> > r - rho resistivity for copper is 0.000 000 017 ohm-meter
> > 
> > R= 0.000 017 ohm = 0.017 milliohm.
> > 
> > I'm impressed. about a factor of 5 margin of safety.
> > 
> > Now 2 Km by 2m^2 is a volume of 4000 m^3
> > Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> > total weight 35,840,000 kg
> > Now I'm impressed again. Cost is a little over $25 million.
> > 
> > Of course it will be the assembly costs that swamp the material costs.
> > You'd likely want the rails to be continuous. so you'd want to caste the
> > rails in place maybe?Or can it be multiple parallel sheets of copper?
> > (how big is a roll of copper 1km long?)
> > 
> > Physically doable, though challenging. that the navy gave up on their
> > railgun, suggests there are other challenges we haven't identified.
> > 
> > Nothing better be around the launch tip, that is going to be
> > one hell of a sonic boom!
> > 
> > Ed
> 
> I think the copper is going to melt the second you make a single firing.
> Well, melt maybe is not the right word. Explosively vaporize is perhaps
> better. Fairly simple to calculate the heating load and then ask whether
> copper’s thermal conductivity is anywhere near what it would need to be. 
> 
> Quenching of accelerator superconducting dipoles is an object lesson. 
> 
> -- 
> Odd Bodkin — Maker of fine toys, tools, tables


The calculations for the required size of the current carrying
rails seem right. The problem is maintaining a firm contact
between the rails and the payload. That part I did skip.
Based on a couple of those railgun videos that seems the
point where vaporization will occur. It seems solvable.


But in thinking this through, I do think there is another
problem involved. Each launch is going to generate one
hell of an EMP (ElectroMagnetic Pulse).

How close can you be before you cell phone gets fried?

Ed

[toc] | [prev] | [next] | [standalone]

#691292

On Wednesday, 20 June 2018 03:39:09 UTC+10, Edward Prochak  wrote:
> On Monday, June 18, 2018 at 5:02:57 PM UTC-4, Odd Bodkin wrote:
> > Edward Prochak <edprochak@gmail.com> wrote:
> > > On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> > >> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> > >>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> > >>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> > >>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > >>>>>> Did 3200 amps with own two brown hands
> > >>>>> 
> > >>>>> you are off by 3 orders of magnitudes!
> > >>>> 
> > >>>> No.
> > >>>> 
> > >>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> > >>>> circuit resistance to get 3200 amps on the average.  The 3200 A
> > >>>> was calculated from the drop in the voltage across the capacitor
> > >>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> > >>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> > >>>> 
> > >>>> Agreed?
> > >>> 
> > >>> Yes, but
> > >>> That is still 3 orders of magnitude less than the
> > >>> 25,000,000 Amps needed for the 1km rail gun.
> > >>> I don't know what materials will carry that high a current.
> > >>> 
> > >>> ED
> > >> 
> > >> In my PhD work, my guide asked me to do a worksheet showing
> > >> how much a conductor of a given dimension will heat up over
> > >> time with current.
> > >> 
> > >> That I did, and with that help I designed my 3200A gun. It does not
> > >> become hot by the way, thanks to the dimensioning.
> > >> 
> > >> Extending that to my motor (the IFE) I worked out that with a
> > >> 500mm*500mm copper cross section, the heat increase would be manageable
> > >> for 3 million amperes.
> > >> 
> > >> I can use the same worksheet to work out the dimensions for 25million amps.
> > >> 
> > >> My guess is that it would be around 2000mm*1000mm.
> > >> 
> > >> That is a lot of copper, for 2000m rails, but it will carry 25 million
> > >> amps. Note, this are rough figures. We also need to know that the whole
> > >> barrel will be vacuum sealed and as cold as possible. Colder the
> > >> temperature, better the conductivity, lower the contact resistance, etc.
> > >> I am not talking about superconductivity, just higher conductivity with lower temperature.
> > >> 
> > >> I am giving over-the-top back of the envelope estimates. Naturally, when
> > >> we get down to the calculas and do the fine slicing, the actual figures
> > >> for a given design will emerge. Reducing the contact resistance is the key. 
> > >> 
> > >> Another point - we need existing defunct mineshafts going down 1-2Km for
> > >> this gun. No point creating a 1000 m vertical structure, what! Natural
> > >> side-support, in a hole!
> > >> 
> > >> I think we can with a limited budget make a gun that can launch small
> > >> satellites into space. It can compete with rocket launches. 
> > >> 
> > >> Cheers,
> > >> Arindam Banerjee
> > > 
> > > Earlier you Arindam) posted:
> > > "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> > > No problem at all, both in theory and practice."
> > > 
> > > the total resistance needs to be 1 milliohm (or less)
> > > 
> > > So what is the resistance?
> > > to obtain a voltage drop of 25000V over 2 kilometers,
> > > R=rl/A
> > > l - length 2 km (=2000 m)
> > > A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> > > r - rho resistivity for copper is 0.000 000 017 ohm-meter
> > > 
> > > R= 0.000 017 ohm = 0.017 milliohm.
> > > 
> > > I'm impressed. about a factor of 5 margin of safety.
> > > 
> > > Now 2 Km by 2m^2 is a volume of 4000 m^3
> > > Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> > > total weight 35,840,000 kg
> > > Now I'm impressed again. Cost is a little over $25 million.
> > > 
> > > Of course it will be the assembly costs that swamp the material costs.
> > > You'd likely want the rails to be continuous. so you'd want to caste the
> > > rails in place maybe?Or can it be multiple parallel sheets of copper?
> > > (how big is a roll of copper 1km long?)
> > > 
> > > Physically doable, though challenging. that the navy gave up on their
> > > railgun, suggests there are other challenges we haven't identified.
> > > 
> > > Nothing better be around the launch tip, that is going to be
> > > one hell of a sonic boom!
> > > 
> > > Ed
> > 
> > I think the copper is going to melt the second you make a single firing.
> > Well, melt maybe is not the right word. Explosively vaporize is perhaps
> > better. Fairly simple to calculate the heating load and then ask whether
> > copper’s thermal conductivity is anywhere near what it would need to be. 
> > 
> > Quenching of accelerator superconducting dipoles is an object lesson. 
> > 
> > -- 
> > Odd Bodkin — Maker of fine toys, tools, tables
> 
> 
> The calculations for the required size of the current carrying
> rails seem right. The problem is maintaining a firm contact
> between the rails and the payload. That part I did skip.
> Based on a couple of those railgun videos that seems the
> point where vaporization will occur. It seems solvable.

I have ideas for the low contact resistance (the most crucial part of the design in practical terms) but let us say they belong to the category of industrial secrets. Already I have given out some: vacuum and cooling, but there are other electrical and mechanical matters for the very low resistance required. 

> 
> 
> But in thinking this through, I do think there is another
> problem involved. Each launch is going to generate one
> hell of an EMP (ElectroMagnetic Pulse).

That was what my guide also thought would happen. I had no idea about what would happen when I suddenly created a huge current. That as you see happened when a thick armature dropped on rails, drawing on the average 2500 amps going up to 3200 amps depending upon a number of factors. Nothing adverse happened. However one wonderful result was noted - after the capacitor bank discharged from say 16V to 8V, it also got charged up a little on the open circuit. You may see it on the video film. The voltmeter shows how it is getting charged up from an open circuit. Evidently the electro-magnetic field strength in the aether medium is charging up the capacitor bank - this is a clear indication of the existence of aether, which is absorbing the energy and then releasing it back. So definitely there will be em energy in aether when the rail gun armature leaves the gun. How much that will impact? I don't think it will be harmful, nothing like a nuclear explosion at any rate.
> 
> How close can you be before you cell phone gets fried?
> 
> Ed

[toc] | [prev] | [next] | [standalone]

#691332

On Wednesday, 20 June 2018 08:35:14 UTC+10, Arindam Banerjee  wrote:
> On Wednesday, 20 June 2018 03:39:09 UTC+10, Edward Prochak  wrote:
> > On Monday, June 18, 2018 at 5:02:57 PM UTC-4, Odd Bodkin wrote:
> > > Edward Prochak <edprochak@gmail.com> wrote:
> > > > On Sunday, June 17, 2018 at 10:07:28 PM UTC-4, Arindam Banerjee wrote:
> > > >> On Monday, June 18, 2018 at 11:44:42 AM UTC+10, Edward Prochak wrote:
> > > >>> On Sunday, June 17, 2018 at 9:13:39 PM UTC-4, Arindam Banerjee wrote:
> > > >>>> On Monday, June 18, 2018 at 10:04:56 AM UTC+10, Edward Prochak wrote:
> > > >>>>> On Friday, June 15, 2018 at 9:21:25 PM UTC-4, Arindam Banerjee wrote:
> > > >>>>>> Did 3200 amps with own two brown hands
> > > >>>>> 
> > > >>>>> you are off by 3 orders of magnitudes!
> > > >>>> 
> > > >>>> No.
> > > >>>> 
> > > >>>> I shorted with a 16.5 V 250F capacitor bank with a low milliohm
> > > >>>> circuit resistance to get 3200 amps on the average.  The 3200 A
> > > >>>> was calculated from the drop in the voltage across the capacitor
> > > >>>> bank. Say when it dropped from 16.5 V to 7.5 V then the charge
> > > >>>> discharged is 250 (16.5 - 7.5) = 2250 coulombs.
> > > >>>> 
> > > >>>> Agreed?
> > > >>> 
> > > >>> Yes, but
> > > >>> That is still 3 orders of magnitude less than the
> > > >>> 25,000,000 Amps needed for the 1km rail gun.
> > > >>> I don't know what materials will carry that high a current.
> > > >>> 
> > > >>> ED
> > > >> 
> > > >> In my PhD work, my guide asked me to do a worksheet showing
> > > >> how much a conductor of a given dimension will heat up over
> > > >> time with current.
> > > >> 
> > > >> That I did, and with that help I designed my 3200A gun. It does not
> > > >> become hot by the way, thanks to the dimensioning.
> > > >> 
> > > >> Extending that to my motor (the IFE) I worked out that with a
> > > >> 500mm*500mm copper cross section, the heat increase would be manageable
> > > >> for 3 million amperes.
> > > >> 
> > > >> I can use the same worksheet to work out the dimensions for 25million amps.
> > > >> 
> > > >> My guess is that it would be around 2000mm*1000mm.
> > > >> 
> > > >> That is a lot of copper, for 2000m rails, but it will carry 25 million
> > > >> amps. Note, this are rough figures. We also need to know that the whole
> > > >> barrel will be vacuum sealed and as cold as possible. Colder the
> > > >> temperature, better the conductivity, lower the contact resistance, etc.
> > > >> I am not talking about superconductivity, just higher conductivity with lower temperature.
> > > >> 
> > > >> I am giving over-the-top back of the envelope estimates. Naturally, when
> > > >> we get down to the calculas and do the fine slicing, the actual figures
> > > >> for a given design will emerge. Reducing the contact resistance is the key. 
> > > >> 
> > > >> Another point - we need existing defunct mineshafts going down 1-2Km for
> > > >> this gun. No point creating a 1000 m vertical structure, what! Natural
> > > >> side-support, in a hole!
> > > >> 
> > > >> I think we can with a limited budget make a gun that can launch small
> > > >> satellites into space. It can compete with rocket launches. 
> > > >> 
> > > >> Cheers,
> > > >> Arindam Banerjee
> > > > 
> > > > Earlier you Arindam) posted:
> > > > "25 million amps is nothing but 25000 volts divided by 1 milliohm. 
> > > > No problem at all, both in theory and practice."
> > > > 
> > > > the total resistance needs to be 1 milliohm (or less)
> > > > 
> > > > So what is the resistance?
> > > > to obtain a voltage drop of 25000V over 2 kilometers,
> > > > R=rl/A
> > > > l - length 2 km (=2000 m)
> > > > A - cross-sectional area 2000mm by 1000mm (2m*1m= 2m^2)
> > > > r - rho resistivity for copper is 0.000 000 017 ohm-meter
> > > > 
> > > > R= 0.000 017 ohm = 0.017 milliohm.
> > > > 
> > > > I'm impressed. about a factor of 5 margin of safety.
> > > > 
> > > > Now 2 Km by 2m^2 is a volume of 4000 m^3
> > > > Density is 8.96 g/cm^3 * 1kg/1000g * (100 cm)^3/m^3 =8960 kg/m^3
> > > > total weight 35,840,000 kg
> > > > Now I'm impressed again. Cost is a little over $25 million.
> > > > 
> > > > Of course it will be the assembly costs that swamp the material costs.
> > > > You'd likely want the rails to be continuous. so you'd want to caste the
> > > > rails in place maybe?Or can it be multiple parallel sheets of copper?
> > > > (how big is a roll of copper 1km long?)
> > > > 
> > > > Physically doable, though challenging. that the navy gave up on their
> > > > railgun, suggests there are other challenges we haven't identified.
> > > > 
> > > > Nothing better be around the launch tip, that is going to be
> > > > one hell of a sonic boom!
> > > > 
> > > > Ed
> > > 
> > > I think the copper is going to melt the second you make a single firing.
> > > Well, melt maybe is not the right word. Explosively vaporize is perhaps
> > > better. Fairly simple to calculate the heating load and then ask whether
> > > copper’s thermal conductivity is anywhere near what it would need to be. 
> > > 
> > > Quenching of accelerator superconducting dipoles is an object lesson. 
> > > 
> > > -- 
> > > Odd Bodkin — Maker of fine toys, tools, tables
> > 
> > 
> > The calculations for the required size of the current carrying
> > rails seem right. The problem is maintaining a firm contact
> > between the rails and the payload. That part I did skip.
> > Based on a couple of those railgun videos that seems the
> > point where vaporization will occur. It seems solvable.
> 
> I have ideas for the low contact resistance (the most crucial part of the design in practical terms) but let us say they belong to the category of industrial secrets. Already I have given out some: vacuum and cooling, but there are other electrical and mechanical matters for the very low resistance required. 


The main issue in practical rail guns for warfare has always been the high current through the armature, causing plasma of the conductor around the sabot or bullet (a hard plastic). The millions of amperes melts the thin metal or rather boils it, becoming a conducting plasma with unfortunately a high resistance needing high voltage... To get around this for my experiments, I created a rail gun which would not need the high voltage for there would be no plasma formed. What did happen was that heavy armature (a brass cylinder) got pock marked with the high current, and so would not last much experimentation! So I wrapped it with aluminium foil for the experiments that are shown in my video films. The current burnt through the foils but did not harm the brass cylinder as had been the case for the very first try.

A similar wrapping technique (around the edges of the 400Kg projectile and over the large rails) could well be done for the 25 million amp current. It is a huge current, but won't last long, what! The 400Kg projectile mass should have enough in it to withstand the IIR heat, all depends upon the design. How low is the 400Kg resistance? Depends on the geometry. If say it is 1 microohm, then the heating loss will be 6.25E8 watts or say 1.5E8 calories second. The heat will last for 0.036 seconds, so say 6E6 calories will be expended in 400Kg or 4*E5 grams. Assuming specific heat of say aluminium (used for the projectile) as 0.2 the the rise of temperature will be 6000000/4*E5*.2= 60/.8 = 75 degrees.

Which is very good! Aluminum melts at over 600deg so 75 deg rise in temperature is acceptable. Even if the current flows through 100Kg of the projectile, still the rise will be 300 deg which is fine especially if the armature is cooled way below zero deg to begin with.

In other words, sending mass to orbit Earth, or Moon or Mars, is perfectly possible with my new design rail gun.  However, I did not do the PhD work for this! I did it to show that there is no reaction to rail gun action, as evident from my experiments. This will revise the whole of physics: entropy, relativity, quantum theory will be irrelevant and aether will come back with electromagnetic theory to explain the workings of Nature. Further, the rail gun will be modified to an extremely fast and efficient motor for space and air travel. 

The sooner the persecution I endure from the whole world ends, the faster the aove most necessary scientific breakthroughs and technical innovations can be realised for the good of all that is animal, mineral and vegetable for all the succeeding centuries.

Cheers,
Arindam Banerjee
> 
> > 
> > 
> > But in thinking this through, I do think there is another
> > problem involved. Each launch is going to generate one
> > hell of an EMP (ElectroMagnetic Pulse).
> 
> That was what my guide also thought would happen. I had no idea about what would happen when I suddenly created a huge current. That as you see happened when a thick armature dropped on rails, drawing on the average 2500 amps going up to 3200 amps depending upon a number of factors. Nothing adverse happened. However one wonderful result was noted - after the capacitor bank discharged from say 16V to 8V, it also got charged up a little on the open circuit. You may see it on the video film. The voltmeter shows how it is getting charged up from an open circuit. Evidently the electro-magnetic field strength in the aether medium is charging up the capacitor bank - this is a clear indication of the existence of aether, which is absorbing the energy and then releasing it back. So definitely there will be em energy in aether when the rail gun armature leaves the gun. How much that will impact? I don't think it will be harmful, nothing like a nuclear explosion at any rate.
> > 
> > How close can you be before you cell phone gets fried?
> > 
> > Ed

[toc] | [prev] | [next] | [standalone]

#691380

On Tuesday, June 19, 2018 at 6:35:14 PM UTC-4, Arindam Banerjee wrote:
> On Wednesday, 20 June 2018 03:39:09 UTC+10, Edward Prochak  wrote:

> > 
> > But in thinking this through, I do think there is another
> > problem involved. Each launch is going to generate one
> > hell of an EMP (ElectroMagnetic Pulse).
> 
> That was what my guide also thought would happen. I had no idea
> about what would happen when I suddenly created a huge current.
> That as you see happened when a thick armature dropped on rails,
> drawing on the average 2500 amps going up to 3200 amps depending
> upon a number of factors. Nothing adverse happened.

Are you sure?

> However one wonderful result was noted - after the capacitor
> bank discharged from say 16V to 8V, it also got charged up a
> little on the open circuit. You may see it on the video film.
> The voltmeter shows how it is getting charged up from an open
> circuit. Evidently the electro-magnetic field strength in the
> aether medium is charging up the capacitor bank
> - this is a clear indication of the existence of aether, which
> is absorbing the energy and then releasing it back.

Now you are reaching for evidence.

> So definitely there will be em energy in aether when the rail
> gun armature leaves the gun. How much that will impact? I don't
> think it will be harmful, nothing like a nuclear explosion at
> any rate.

Agree, this is not nuclear level EMP, but a 25 million amp
circuit over 2 KM. (How big is that Magnetic field?)
So I am still curious for an answer to my question:

> > How close can you be before you cell phone gets fried?
> > 
> > Ed

[toc] | [prev] | [next] | [standalone]

#691406

On Thursday, June 21, 2018 at 3:49:35 AM UTC+10, Edward Prochak wrote:
> On Tuesday, June 19, 2018 at 6:35:14 PM UTC-4, Arindam Banerjee wrote:
> > On Wednesday, 20 June 2018 03:39:09 UTC+10, Edward Prochak  wrote:
> 
> > > 
> > > But in thinking this through, I do think there is another
> > > problem involved. Each launch is going to generate one
> > > hell of an EMP (ElectroMagnetic Pulse).
> > 
> > That was what my guide also thought would happen. I had no idea
> > about what would happen when I suddenly created a huge current.
> > That as you see happened when a thick armature dropped on rails,
> > drawing on the average 2500 amps going up to 3200 amps depending
> > upon a number of factors. Nothing adverse happened.
> 
> Are you sure?

Yes. I did many experiments generating 2500A within milliseconds. The first time I did not know what would happen. Nothing happened to anything electronic around, like the meter. 

EMP happens with a nuclear explosion. That is a lot more energy and that too realised far more quickly to really shake up the aether. Huge voltage spikes result from such vibration and they can destroy electronics equipment easily.
 
> > However one wonderful result was noted - after the capacitor
> > bank discharged from say 16V to 8V, it also got charged up a
> > little on the open circuit. You may see it on the video film.
> > The voltmeter shows how it is getting charged up from an open
> > circuit. Evidently the electro-magnetic field strength in the
> > aether medium is charging up the capacitor bank
> > - this is a clear indication of the existence of aether, which
> > is absorbing the energy and then releasing it back.
> 
> Now you are reaching for evidence.


Anyone can see from my videos how the capacitor bank charges up on an open circuit. There has to be some sort of mass around to do that, by retaining the energy first and then sending it back to the source. A pretty conclusive evidence for aether - and this at the direct current/field level, as opposed to the alternative fields of electromagnetic wave propagation.


Should be 1 NPPhy for me here, but never mind.
> 
> > So definitely there will be em energy in aether when the rail
> > gun armature leaves the gun. How much that will impact? I don't
> > think it will be harmful, nothing like a nuclear explosion at
> > any rate.
> 
> Agree, this is not nuclear level EMP, but a 25 million amp
> circuit over 2 KM. (How big is that Magnetic field?)
> So I am still curious for an answer to my question:

Whatever big that is, it will be contained locally, affecting mostly the armature. Biot-Savart law will show pretty clearly how much the magnitude of the magnetic field will be at any distance. It is a inverse distance law relation. https://en.wikipedia.org/wiki/Magnetic_field  tells us:

B=muI/(2*pi*r)
B=4piE-7*25E6/(2*pi*r)
B=5/r where r is in meters.

Don't think there is much to worry, what.

 I have no doubt there will be thus a fairly strong magnetic field generated for 0.036seconds, so proper precautions need to be taken for those wearing say pacemakers or other electronic implants within say 20 meters.
> 
> > > How close can you be before you cell phone gets fried?


> > > 
> > > Ed

[toc] | [prev] | [next] | [standalone]

#691482

On Wednesday, June 20, 2018 at 7:18:25 PM UTC-4, Arindam Banerjee wrote:
> On Thursday, June 21, 2018 at 3:49:35 AM UTC+10, Edward Prochak wrote:
> > 
> > Agree, this is not nuclear level EMP, but a 25 million amp
> > circuit over 2 KM. (How big is that Magnetic field?)
> > So I am still curious for an answer to my question:
> 
> Whatever big that is, it will be contained locally,
> affecting mostly the armature. Biot-Savart law will
> show pretty clearly how much the magnitude of the
> magnetic field will be at any distance. It is a
> inverse distance law relation.
> https://en.wikipedia.org/wiki/Magnetic_field
>  tells us:
> 
> B=muI/(2*pi*r)
> B=4piE-7*25E6/(2*pi*r)
> B=5/r where r is in meters.
> 
> Don't think there is much to worry, what.
> 
>  I have no doubt there will be thus a fairly strong
> magnetic field generated for 0.036seconds, so proper
> precautions need to be taken for those wearing say
> pacemakers or other electronic implants within say 20 meters.

Really that short a distance? I trust you Arindam.
But I may check it offline anyway 8^)

Ed

[toc] | [prev] | [next] | [standalone]

Page 1 of 3  [1] 2 3  Next page →

Back to top | Article view | sci.physics

csiph-web